Chemistry - Electrochemistry Question with Solution | TestHub

Half cell 1: ${\mathrm{ClO}}_3^{-}+6{\mathrm{H}}^{+}+6e^{-}\to {\mathrm{Cl}}^{-}+3{\mathrm{H}}_2\mathrm{O}$.

$E_1=1.45-\frac{0.06}{6}\log \frac{\left[{\mathrm{Cl}}^{-}\right]}{\left[{\mathrm{ClO}}_3^{-}\right]{\left[{\mathrm{H}}^{+}\right]}^6}=1.45-0.01\log \frac{0.1}{0.2\times (0.5{)}^6}$. The ratio is $\frac{0.1}{0.003125}=32$, $\log 32=1.5$, so $E_1=1.45-0.015=1.435$ V.

Half cell 2: ${\mathrm{IO}}_3^{-}+6{\mathrm{H}}^{+}+6e^{-}\to {\mathrm{I}}^{-}+3{\mathrm{H}}_2\mathrm{O}$.

$E_2=1.08-0.01\log \frac{0.4}{0.1\times (0.1{)}^6}$. The ratio is $4\times 10^6$, $\log =6.6$, so $E_2=1.08-0.066=1.014$ V.

Half cell 1 has the higher potential and is the cathode.

EMF $=1.435-1.014=0.42$ V.

Answer: 0.42 V.